Question: $v(t) = (\cos(t)\sin(t), \sin(t)\sin(t), \cos(t))$ What is the velocity of $v(t)$ ? Choose 1 answer: Choose 1 answer: (Choice A) A $(1 + 2\sin^2(t), \sin(2t), -\sin(t))$ (Choice B) B $(1 + 2\sin^2(t), \cos(2t), -\sin(t))$ (Choice C) C $(1 - 2\sin^2(t), \cos(2t), -\sin(t))$ (Choice D) D $(1 - 2\sin^2(t), \sin(2t), -\sin(t))$
Solution: The velocity of a parametric curve is the derivative of its position. If $f(t) = (a(t), b(t), c(t))$, then the velocity is: $f'(t) = (a'(t), b'(t), c'(t))$ Our position function here is $v(t)$. $\begin{aligned} v'(t) &= (-\sin^2(t) + \cos^2(t), 2\sin(t)\cos(t), -\sin(t)) \\ \\ &= (1 - 2\sin^2(t), \sin(2t), -\sin(t)) \end{aligned}$ Therefore, the velocity of $v(t)$ is: $(1 - 2\sin^2(t), \sin(2t), -\sin(t))$